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As I work on certain problems that are relevant to me, I come across questions that make me realize how much just knowing the basic apparatus of results is not enough..

To simplify as much as possible. Let's say that we have a sequence of measures (not necessarily probability measures, but positive and one can even add the requirement that $\inf_n \mu_n(R)> 0 $ if need be) $\{\mu_n: n \in N\} $ that are known (proved already) to converge weakly to a finite measure $\mu. $ Then we know that if $f $ is a bounded continuous function, $$ \int f(x)\, \mu_n(dx) \to \int f(x)\, \mu(dx) $$ as $n \to \infty. $ What happens to terms like, say, $$ \int_{|x|\le c} f(x)\, \mu_n(dx) $$ and/or $$ \int_{|x| > c} f(x)\, \mu_n(dx)~? $$

Also, is the choice of the points $\pm c $ completely free or is it necessary to assume more structure like requiring that $c $ be a point of continuity for $\mu, $ i.e., $\mu(\{\pm c\}) = 0? $ In this case it would look like that if we consider the integral $$ \int_{|x|\le c} f(x)\, \mu_n(dx) $$ then it would be possible to use the fact that $$\biggl| \int_{|x|\le c} f(x)\, \mu_n(dx) - \int_{|x|\le c} f(x)\, \mu(dx)\biggl| \le K |\mu_n([-c, c]) - \mu([-c,c])| \to 0 $$ using the definition of weak convergence and the fact that $\mu(\partial [-c, c]) = 0 $ where $\partial $ denotes the boundary of $[-c, c]. $

And, if the requirement that $c $ be indeed a point of continuity for $\mu $ were necessary, is there any easy counter-example to the fact that we cannot choose the points $\pm c $ completely arbitrary?

If anyone has any wisdom to share on these issues, I would be deeply grateful.

Maurice

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Consider e.g. $\mu_n(dx) = (\delta(x - 1 - 1/n) - \delta(x - 1 + 1/n))\; dx$ (i.e. a unit mass at $1+1/n$ minus a unit mass at $1-1/n$), which converges weakly to $\mu = 0$. But $$\int_{|x| > 1} f(x)\; \mu_n(dx) = \int_{|x| \ge 1} f(x)\; \mu_n(dx) = f(1 + 1/n) \to f(1)$$ So continuity of $\mu$ is not enough.

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  • $\begingroup$ Thank you Robert. In your counter-example, unless I am making a silly mistake, the measure you suggested would be such that $\mu_n(\{1-1/n\})=-1 $ and hence $\mu_n $ would not be a measure. $\endgroup$
    – Maurice
    Dec 1, 2014 at 19:19
  • $\begingroup$ It's not a positive measure. You didn't say you wanted positive measures. $\endgroup$ Dec 1, 2014 at 19:44
  • $\begingroup$ I apologize for that.. Yes, I am working with measures. Signed measures might be interesting on its own merit, but at this stage I would be happy if I could solve my problems with "normal" measures. $\endgroup$
    – Maurice
    Dec 1, 2014 at 19:53

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